What is the integral of 1/(P(1-P/K)) ?

The integral of 1/(P(1-P/K)) is ln((|P|)/(|K|))-ln| P/K-1|+C

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integral of 1/(P(1-P/K))

\int \frac{1}{P ( 1 - \frac{P}{K} )} d P

ln (\frac{∣ P ∣}{∣ K ∣}) - ln \frac{P}{K} - 1 + C

Solution steps

\int \frac{1}{P ( 1 - \frac{P}{K} )} d P

Factor

\frac{1}{( 1 - \frac{P}{K} ) P}

= \int \frac{1}{- P ( \frac{P}{K} - 1 )} d P

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{P ( \frac{P}{K} - 1 )} d P

Apply u-substitution

: \int \frac{1}{u ( u - 1 )} d u

= - \int \frac{1}{u ( u - 1 )} d u

Take the partial fraction of

\frac{1}{u ( u - 1 )} : - \frac{1}{u} + \frac{1}{u - 1}

= - \int - \frac{1}{u} + \frac{1}{u - 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int \frac{1}{u} d u + \int \frac{1}{u - 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= - (- ln ∣ u ∣ + ln ∣ u - 1 ∣)

Substitute back

u = \frac{P}{K}

= - (- ln \frac{P}{K} + ln \frac{P}{K} - 1)

Simplify

- (- ln \frac{P}{K} + ln \frac{P}{K} - 1) : ln (\frac{∣ P ∣}{∣ K ∣}) - ln \frac{P}{K} - 1

= ln (\frac{∣ P ∣}{∣ K ∣}) - ln \frac{P}{K} - 1

Add a constant to the solution

= ln (\frac{∣ P ∣}{∣ K ∣}) - ln \frac{P}{K} - 1 + C

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What is the integral of 1/(P(1-P/K)) ?
The integral of 1/(P(1-P/K)) is ln((|P|)/(|K|))-ln| P/K-1|+C